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edited 115

ur key is wrong no doubt about it .......

i didnt get uuu

y=x(rt(1-x)) now differenciate

$y^I=2(1+x)+x/2(\sqrt{1+x})$

now at (0,0) implie equation is x=y

cheeeeerrrrrrssssss

from above 2eq naturally x-y is common line

hey I got only one way of doing ur question
arbitrarily take a=b=c=1as these values satisfies the condition.
Now substitute in the question u will get x^2-1=0
so the equation has got real roots.

Hey aggmegha dont be sad sriram is there know.

apply transformation

2sin3xcos2x=sin3x

sin3x(2cos2x-1)=0

sin3x=0 cos2x=1/2

3x=n $\pi$ 2x=2n $\pi$ +- $\pi$ /3

x=0, $\pi$ /3,2 $\pi$ /3, $\pi$ x= $\pi$ /6,5 $\pi$ /6

take x=tanA,y=tanB,z=tanC

tanB(tanA+tanC)=1-tanAtanC

tan(A+C)=cotB

$\Sigma$ tan(A+C)=xy+yz+zx/xyz=1/xyz

I got 1/xyz. is it right

Hey vary simple Li has hich M.P.It liberates high energybut not sufficient to melt it.While in case of others their M.P is low and the energy is sufficient to melt it.Hence the metal gets exposed to larger surface area.So other elements of 1A acts vigorously with water.

yaar simple Intermediate question......

a=2RsinA,b=2RsinB,r1=2RsinA/2cosB/2cosC/2,r2=2RcosA/2sinB/2cosC/2,r3=2RcosA/2cosB/2sinC/2

Substitute and transform.Simple

arsy yaar very simple just diferenciate both the equation and multiply them you will get -1

I got the answer as x+1/x-1.Is it right.if it is right i will give u explanation

Now rate me for both the above questions

31)yaar simple take a charge at perpendicular bisector at a distance x.

now resolve the components.The forces perpendicular to bisector cancels.

$F=\frac{kQqx}{\sqrt{(d^2/4+x^2})^3}$ .

now for force to be maximum $\frac{dF}{dx}=0$ .

solve it x= $\frac{d}{2\sqrt{2}}$ .now substitute in F to get maximum force maximum.